# Help with hard equation

1. Nov 22, 2005

### symplectic_manifold

Hello!
We've got to solve the following equation:
$i \ Re \ z+Im \ \bar{z}+\pi=|z|+arg(z)$ with $arg(z)\in[0,2\pi)$
Please don't solve it for me. Give me a hint first.

Last edited: Nov 22, 2005
2. Nov 22, 2005

### HallsofIvy

Staff Emeritus
First write z= x+ iy= |z|(cos(arg z)+ i sin(arg z))
of course, $|z|= \sqrt{x^2+ y^2}$ and $arg z= arctan(\frac{y}{x})$.
Now you can write the condition in terms of x and y.

Last edited: Nov 22, 2005
3. Nov 22, 2005

### symplectic_manifold

Ok, thanks...

Did you mean that arg(z)=arctan(y/x)?

Now it looks like this doesn't it?:
$ix-y+\pi=\sqrt{x^2+y^2}+arctan\left(\frac{y}{x}\right)$
or $(-y-\sqrt{x^2+y^2})+ix=arctan\left(\frac{y}{x}\right)-\pi$
Now it seems the number on the right is real, so the number on the left must be real too, so x=0, but then arctan(y/x) is not defined...hm...does it have any consequences?

4. Nov 22, 2005

### BerkMath

Sure it does, arctan(inf)=pi/2 the answer is then seen to be z=0+i*(pi/4).

5. Nov 22, 2005

### symplectic_manifold

Ok, thanks guys! :)